Integrand size = 23, antiderivative size = 91 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=-\frac {2 b}{11 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}-\frac {16 b}{77 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}-\frac {64 b}{231 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \] Output:
-2/11*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(11/2)-16/77*b/f/(b*sec(f*x+e))^ (3/2)/sin(f*x+e)^(7/2)-64/231*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(3/2)
Time = 0.78 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\frac {2 b (-45+28 \cos (2 (e+f x))-4 \cos (4 (e+f x)))}{231 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)} \] Input:
Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(13/2)),x]
Output:
(2*b*(-45 + 28*Cos[2*(e + f*x)] - 4*Cos[4*(e + f*x)]))/(231*f*(b*Sec[e + f *x])^(3/2)*Sin[e + f*x]^(11/2))
Time = 0.67 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3064, 3042, 3064, 3042, 3058}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sin ^{\frac {13}{2}}(e+f x) \sqrt {b \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (e+f x)^{13/2} \sqrt {b \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3064 |
\(\displaystyle \frac {8}{11} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)}dx-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{11} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin (e+f x)^{9/2}}dx-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3064 |
\(\displaystyle \frac {8}{11} \left (\frac {4}{7} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {5}{2}}(e+f x)}dx-\frac {2 b}{7 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{11} \left (\frac {4}{7} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin (e+f x)^{5/2}}dx-\frac {2 b}{7 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3058 |
\(\displaystyle \frac {8}{11} \left (-\frac {8 b}{21 f \sin ^{\frac {3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {2 b}{7 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\) |
Input:
Int[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(13/2)),x]
Output:
(8*((-2*b)/(7*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7/2)) - (8*b)/(21*f*( b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(3/2))))/11 - (2*b)/(11*f*(b*Sec[e + f* x])^(3/2)*Sin[e + f*x]^(11/2))
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^( m_.), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1 )/(a*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m - n + 2, 0] & & NeQ[m, -1]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/ (a*f*(m + 1))), x] + Simp[(m - n + 2)/(a^2*(m + 1)) Int[(a*Sin[e + f*x])^ (m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, - 1] && IntegersQ[2*m, 2*n]
Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58
method | result | size |
default | \(-\frac {2 \left (32 \cos \left (f x +e \right )^{5}-88 \cos \left (f x +e \right )^{3}+77 \cos \left (f x +e \right )\right )}{231 f \sin \left (f x +e \right )^{\frac {11}{2}} \sqrt {b \sec \left (f x +e \right )}}\) | \(53\) |
Input:
int(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(13/2),x,method=_RETURNVERBOSE)
Output:
-2/231/f/sin(f*x+e)^(11/2)/(b*sec(f*x+e))^(1/2)*(32*cos(f*x+e)^5-88*cos(f* x+e)^3+77*cos(f*x+e))
Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\frac {2 \, {\left (32 \, \cos \left (f x + e\right )^{6} - 88 \, \cos \left (f x + e\right )^{4} + 77 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{231 \, {\left (b f \cos \left (f x + e\right )^{6} - 3 \, b f \cos \left (f x + e\right )^{4} + 3 \, b f \cos \left (f x + e\right )^{2} - b f\right )}} \] Input:
integrate(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(13/2),x, algorithm="fricas")
Output:
2/231*(32*cos(f*x + e)^6 - 88*cos(f*x + e)^4 + 77*cos(f*x + e)^2)*sqrt(b/c os(f*x + e))*sqrt(sin(f*x + e))/(b*f*cos(f*x + e)^6 - 3*b*f*cos(f*x + e)^4 + 3*b*f*cos(f*x + e)^2 - b*f)
Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\text {Timed out} \] Input:
integrate(1/(b*sec(f*x+e))**(1/2)/sin(f*x+e)**(13/2),x)
Output:
Timed out
\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\int { \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {13}{2}}} \,d x } \] Input:
integrate(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(13/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(13/2)), x)
Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\text {Timed out} \] Input:
integrate(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(13/2),x, algorithm="giac")
Output:
Timed out
Time = 33.28 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.79 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\frac {{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\sqrt {\frac {b}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,992{}\mathrm {i}}{231\,b\,f}+\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,608{}\mathrm {i}}{231\,b\,f}-\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,320{}\mathrm {i}}{231\,b\,f}+\frac {{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,64{}\mathrm {i}}{231\,b\,f}\right )\,1{}\mathrm {i}}{32\,{\sin \left (e+f\,x\right )}^{11/2}} \] Input:
int(1/(sin(e + f*x)^(13/2)*(b/cos(e + f*x))^(1/2)),x)
Output:
(exp(- e*6i - f*x*6i)*(b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^ (1/2)*((exp(e*6i + f*x*6i)*992i)/(231*b*f) + (exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*608i)/(231*b*f) - (exp(e*6i + f*x*6i)*cos(4*e + 4*f*x)*320i)/(231* b*f) + (exp(e*6i + f*x*6i)*cos(6*e + 6*f*x)*64i)/(231*b*f))*1i)/(32*sin(e + f*x)^(11/2))
\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right ) \sin \left (f x +e \right )^{7}}d x \right )}{b} \] Input:
int(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(13/2),x)
Output:
(sqrt(b)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x)))/(sec(e + f*x)*sin(e + f*x)**7),x))/b